<think>
Okay, let's tackle this problem step by step. So, the task is to find the maximum straight-line distance from the base camp after three displacements. The options given range from a few meters up to around 5.6 km, but since all the displacements are in kilometers except maybe some options, I need to check that.
First, let me understand what each vector represents:
- Vector a is 2.0 km directly east.
- Vector b is 2.0 km at 30 degrees north of east. That means it has components both east and north.
- Vector c is 1.0 km directly west. But we can also use -c, which would be 1.0 km east.
So, we can choose any combination of these vectors, possibly using their negatives. Wait, the problem says we can substitute either -b or -c for b or c. Hmm. Let me re-read that part.
The problem states: "you may substitute either −⃗b for ⃗b or −⃗c for ⃗c." Oh right! So instead of using vector b (which is 2.0 km at 30° N of E), we could use -b, meaning 2.0 km at 30° S of E. Similarly, instead of using vector c (due west), we can use -c, which is due east. Therefore, we have flexibility in choosing directions for those two vectors.
Our job is to arrange the three displacements (using any permutation of a, b, c with possible substitutions) such that the resultant vector has the maximum magnitude.
Since the problem allows us to pick any order and substitution, our strategy is to maximize the total displacement. To do that, perhaps we want to add vectors in a way that they reinforce each other rather than cancel out.
Let me first consider the individual vectors' components.
Vector a: Eastward component = +2.0 km; North component = 0.
Vector b: 2.0 km at 30° N of E. Its components would be:
East: |b| * cos(θ) = 2.0 * cos(30°)
North: |b| * sin(θ) = 2.0 * sin(30°)
Similarly, vector -b would be 2.0 km at 30° S of E, so its components would be:
East: same as above, 2.0*cos(30°)
South: 2.0*sin(30°). Which would make the north component negative.
Vector c: West is -x direction. So, -c is east. So vector c is (-1.0 km, 0), while -c is (+1.0 km, 0).
Now, the key here is to combine these three vectors in different orders, considering whether to take them as original or reversed (i.e., -b or -c), to get the largest possible net displacement.
But wait, there are three moves. Each move must be one of the vectors a, b, c or their negations. However, substituting -b or -c only applies when using b or c. So actually, the available choices are:
Each of the three moves can be:
Either a (east 2.0 km),
Or b (2.0 km 30NofE),
Or c (west 1.0 km),
or -b (same magnitude, 30SofE),
or -c (east 1.0 km).
Wait, no—the problem says you can substitute either -b for b or -c for c. Does that mean that during the three moves, we can replace one occurrence of b with -b, or one occurrence of c with -c?
Alternatively, does that mean that we can choose between using b or -b, and similarly for c? Or that for each of the three steps, we can choose whether to use the original or substituted version?
Actually, the wording is: "you may use the following displacements in any order: ... [the three vectors] ... Alternatively, you may substitute either −⃗b for ⃗b or −⃗c for ⃗c."
This suggests that for each displacement, you can choose to use either the original vector or its opposite. For instance, for the second move, instead of taking vector b, you can take -b. Similarly for c vs -c. Therefore, for each of the three moves, we can decide whether to take a, b, c or their respective opposites. However, the problem might imply that we have three separate displacements, each being chosen among the possibilities. Wait, but how many times can we use each vector? Since the problem says "you may use the following displacements in any order", implying that you can use each once. Because otherwise, if you had unlimited uses, then you'd just keep adding vectors until you reach max. But since there are three moves, probably each displacement is used once—so the three moves consist of selecting one of {a, b, c} or {-a, -b, -c}? No, because the original list includes a, b, c. Then substituting -b or -c implies that when you use b, you can use it or its opposite. So for each move, you can select either the original or the substituted versions of b or c. Wait, maybe the problem is saying that you can use a, b, c, but also substitute either -b or -c into the set. Maybe that's unclear. Alternatively, perhaps the allowed vectors are a, b, c, and also their negatives. But that seems like overcomplicating.
Alternatively, perhaps the problem gives three possible vectors (a, b, c), and you can choose to use either the original or the negative of each. So for example, for the three moves, you can choose to use a, or -a? Wait, no. Looking back:
Original statement: "You may use the following displacements in any order: (a)... (b)... (c)... Alternatively, you may substitute either −⃗b for ⃗b or −⃗c for ⃗c."
Ah, okay. This suggests that normally, you have access to vectors a, b, c. But you can also substitute either -b or -c. So that means, for example, instead of using vector b, you can use -b. Similarly, instead of using c, you can use -c. But what about a? It doesn't mention anything about substituting -a. So, for vector a, you can only use it as is (since there's no alternative mentioned). Therefore, the three moves can include:
Move 1: Either a, or -a? Wait, no—if the substitution is only for b and c. So the three moves can be a, b, c, or alternatively, for b, use -b, or for c, use -c. So for example, the three moves could be a, b, c; or a, -b, c; or a, b, -c; etc. So each move can be selected as follows:
First move: a, or -a? Well, according to the problem statement, the initial list is (a), (b), (c). Then you can substitute either -b or -c. So perhaps you cannot substitute -a. Thus, the three moves are composed of three vectors, each of which is either a, b, c, or for b/c, you can choose to use -b/-c. Therefore, the three moves can be combinations of a, ±b, ±c, but with the constraint that if you use b or c, you can flip their sign.
Therefore, the three moves can be arranged in any order, and each of the three moves can be either a, or b (with optional negative), or c (optional negative). Wait, but that's conflicting. How can we choose the signs? If the problem says that we can substitute either -b or -c for b or c, then perhaps in addition to the normal vectors a, b, c, we can also use -b or -c. Therefore, for each of the three moves, we can choose between a, b, c, -b, -c? Not sure. But maybe the actual approach is that for the three moves, we have three vectors to choose from, each being either a, b, c, or replacing b with -b or c with -c. Therefore, the three moves are made of three distinct vectors, each of which is a, b, or c (possibly replaced by their negatives). So for example, one possibility is a, b, c; another is a, -b, c; another is a, b, -c; another is -a, b, c... Wait, but can we use -a? According to the problem description, it's only allowing substitution of -b or -c. Therefore, unless the problem allows substituting -a, but it doesn't say so. Hence, the three moves are limited to a, b, c, with the option to use -b or -c instead of b or c. Therefore, in effect, the three moves can be any combination of a, ±b, ±c—but keeping in mind that each of the three moves is one of a, b, c, with the ability to reverse b or c.
However, that interpretation may be incorrect. Another way is that the three displacements are a, b, c. But you can substitute either -b or -c for b or c. So, for example, you can choose to use a, -b, -c. Or a, b, -c. Or something else. So the three moves are determined by choosing for each of the three positions whether to use the original or the substituted version of b or c. But that's getting complicated.
Alternatively, the problem might allow you to use any of the four vectors: a, b, -b, c, -c. But since the problem mentions three moves, perhaps the three moves are selected from a, b, c, and you can choose to swap b with -b or c with -c. But again, without more clarity, perhaps the safest assumption is that the three moves are three vectors, each being a, b, or c, but with the freedom to choose whether to use b or -b, and/or c or -c. Therefore, for example, the three moves could be a, b, c, or a, -b, -c, etc.
In any case, regardless of the exact permutations, the idea is to find the maximum possible magnitude of the sum of the three vectors. Given that, we need to figure out the best combination of vectors that adds up to the maximum possible length.
To approach this systematically, perhaps I should compute the maximum possible displacement by trying different combinations of vectors.
Note that the problem asks for the maximum straight-line distance, i.e., the magnitude of the resultant vector. We need to maximize ||a + b + c||, where each term can be positive or negative depending on whether we use the original or substituted vector.
But since the three moves can be ordered in any sequence, the order shouldn't matter for the result, as vector addition is commutative. So the order of the moves won't affect the final position. Therefore, we can focus on combining the vectors in terms of their components, regardless of order.
Thus, the key variables are the choice of signs for vectors b and c (if needed). Also note that vector a is fixed as +2.0 km east, but we can't change it. Unless the problem allows substituting -a, but based on the problem text, that isn't the case.
Wait, but hold on. Let me read the problem again carefully:
"You may use the following displacements in any order: (a) a, 2.0 km due east; (b) b, 2.0 km 30° north of east; (c) c, 1.0 km due west. Alternatively, you may substitute either −⃗b for ⃗b or −⃗c for ⃗c."
Hmm. The wording suggests that the three displacements are a, b, c. However, you can substitute either -b or -c for b or c. Therefore, for example, if you were going to use vector b, you can instead use -b. Similarly, if you wanted to use c, you can use -c. Therefore, the three moves are formed by choosing one of the three vectors (a, b, c) and then optionally flipping their signs if they're b or c. So, the three moves are:
Move 1: Choose between a, b, or c. If it's b, you can choose to use it or -b. Same for c.
Therefore, the three moves can be represented as:
Optionally, for each move, if it's b, you can use +b or -b. If it's c, you can use +c or -c. And a is always +a.
But since there are three moves, and each can be a, b, c, with possible sign changes, the problem becomes finding the optimal combination of three vectors (each being a, ±b, ±c) such that their sum has maximum magnitude.
But since we can choose the order, but the sum is independent of order, we can proceed to calculate the maximum possible value of the absolute value of the sum of three vectors, where each vector is either a, ±b, or ±c.
But given that there are three moves, and we have to choose three vectors from the set {a, b, c}, each possibly flipped in sign if they are b or c.
Wait, but perhaps the problem allows us to use each of the three vectors once, but with the option to negate b or c. So for example, we can have:
Case 1: Use a, b, c
Case 2: Use a, -b, c
Case 3: Use a, b, -c
Case 4: Use a, -b, -c
Additionally, cases where we don't use a, but maybe use others? Like using b, c, -c? But the problem says "three straight-line moves," so we have to use three moves. But the original three vectors are a, b, c. So perhaps the three moves must be a, b, c in some order, but with the possibility to substitute -b or -c for b or c. Therefore, the three moves will be one of a, plus either b or -b, plus either c or -c. Therefore, the three vectors are a, ±b, ±c. Hence, there are several possibilities:
Possibility 1: All three vectors are positive: a, b, c.
Possibility 2: a, b, -c.
Possibility 3: a, -b, c.
Possibility 4: a, -b, -c.
Also, perhaps other combinations where instead of a, we have -a? But according to the problem, we can't substitute -a. Only -b or -c. Therefore, the three moves must involve a, along with either b or -b and c or -c. So, the possible combinations are those four cases above.
Is that the only way? Are there other combinations where maybe we don't use a? For example, using b, -b, c? But that would require three moves, but the original problem lists a, b, c as the three displacements. So, perhaps you have to use all three: a, b, c, but with the possibility to substitute -b or -c. Therefore, yes, the three moves are a, ±b, ±c.
If that's the case, then we can analyze these four scenarios. Additionally, perhaps there are other combinations where we don't use a, but that would contradict the initial statement of having three moves. But according to the problem statement, the three moves are "you may use the following displacements...", which are a, b, c. Therefore, the three moves must be a, b, c in some order, with possible substitutions of b or c. Therefore, the three vectors are necessarily a, ±b, ±c.
Therefore, now, for each scenario, compute the total displacement vector, then find its magnitude, and see which scenario yields the maximum.
So, let's start analyzing each case.
Case 1: a, b, c. All positives.
Compute the components:
Vector a: (2.0, 0)
Vector b: 2.0 km at 30° N of E → components:
East: 2.0 * cos(30°) ≈ 2.0 * (√3/2) ≈ √3 ≈ 1.732 km
North: 2.0 * sin(30°) = 2.0 * 0.5 = 1.0 km
Vector c: (-1.0, 0) km (West)
Total displacement:
East component: 2.0 + 1.732 - 1.0 = 2.0 + 0.732 = 2.732 km
North component: 0 + 1.0 + 0 = 1.0 km
Magnitude: sqrt((2.732)^2 + (1.0)^2) ≈ sqrt(7.464 + 1) = sqrt(8.464) ≈ 2.91 km
Which is approximately 2.9 km, matching option D.
Case 2: a, b, -c. Here, c is subtracted, so vector -c is (1.0, 0) km.
Vectors:
a: (2.0, 0)
b: same as before: ~1.732, 1.0
-c: (1.0, 0)
Total east: 2.0 + 1.732 + 1.0 = 4.732 km
North: 1.0 km
Magnitude: sqrt(4.732² + 1.0²) ≈ sqrt(22.39 + 1) ≈ sqrt(23.39) ≈ 4.837 km, roughly 4.84 km. Which is close to option H (5.0 km?) Wait, looking at the options:
Options are A: 4.2m, B: 4.6 km, C: 3.8 km, D: 2.9 km, E: 5.2 km, F: 3.5 m, G: 4.8 m, H: 5.0 km, I: 5.6 m, J: 3.2 km.
So 4.837 km would correspond to option G (4.8 m?), but wait, the units in the options are mostly in km except for some that are in meters. However, the problem says that the displacements are in km (for example, 2.0 km, 1.0 km). So the answers likely are in km. Option G is 4.8 m, which is 0.0048 km — clearly wrong. Wait, but the options are written with units, but the problem says "distance". Wait, checking the options again:
Looking at the options:
(A): 4.2 m
(B): 4.6 km
(C): 3.8 km
(D): 2.9 km
(E): 5.2 km
(F): 3.5 m
(G): 4.8 m
(H): 5.0 km
(I): 5.6 m
(J): 3.2 km
So most of the options are in km, except for A, F, G, I which are in meters. However, the given displacements are in km (like 2.0 km, 1.0 km). Therefore, the correct answer should be in km. Therefore, option G is 4.8 m, which is 0.0048 km – very small compared to the rest. Clearly, this is a mistake. Probably, the options meant to have consistent units. Perhaps the options have typos. However, given the context, the answer should be in km. So in Case 2, the magnitude is approx 4.84 km, which is closest to option H (5.0 km)? Or option B (4.6 km)?
Wait, my calculation was approximate. Let's recalculate precisely.
First, precise values:
cos(30°) is √3 / 2 ≈ 0.8660
sin(30°)=0.5
So vector b's components:
East: 2.0 * 0.8660 = 1.732 km
North: 2.0 * 0.5 = 1.0 km
Then, for Case 2: a, b, -c.
East: 2.0 (from a) + 1.732 (from b) + 1.0 (from -c) = 4.732 km
North: 1.0 km (only from b)
So magnitude squared is (4.732)^2 + (1.0)^2.
Calculate 4.732^2:
Well, 4.732 is approximately equal to 4.732. Squaring:
Approximate method: (4.7)^2=22.09, (0.032)^2≈0.001, cross term 2*4.7*0.032≈0.3008. Total approx 22.09+0.3008 + 0.001 ≈ 22.39. Adding 1.0^2=1, total ≈23.39. Square root of 23.39 is approx 4.837 km. Which rounds to 4.84 km.
Looking at the options, B is 4.6 km, H is 5.0 km. So 4.84 is closer to H (but H is 5.0) or maybe none? Wait, perhaps there's a better combination.
Wait, maybe there are other cases. Let's look at the remaining cases.
Case 3: a, -b, c.
Here, vector -b is 2.0 km at 30° south of east. Components:
East: still 2.0 * cos(30°) ≈ 1.732 km
South: 2.0 * sin(30°) = 1.0 km → South component is negative, so north component is -1.0 km.
Vector c is (-1.0, 0)
So total components:
East: 2.0 (from a) + 1.732 (from -b) - 1.0 (from c) = 2.0 + 0.732 = 2.732 km
North: 0 (from a) - 1.0 (from -b) + 0 (from c) = -1.0 km
Magnitude is same as in case 1: sqrt(2.732² + (-1.0)^2) ≈ sqrt(7.464 + 1) = sqrt(8.464) ≈ 2.91 km, same as case 1. So this case gives same magnitude as case 1.
Case 4: a, -b, -c.
Here, -b is as above (components 1.732 east, -1.0 north), and -c is (1.0, 0).
So total components:
East: 2.0 (a) + 1.732 (from -b) + 1.0 (from -c) = 4.732 km
North: 0 (a) -1.0 (from -b) + 0 (from -c) = -1.0 km
So magnitude is sqrt(4.732² + (-1.0)^2) same as case 2, which is ~4.84 km.
So Cases 2 and 4 give similar results (~4.84 km). Now, are there other combinations?
What if we didn't use a? Wait, but according to the problem statement, we have to use the three displacements. Wait, the problem says "you may use the following displacements in any order: (a), (b), (c)", so we have to use all three. Therefore, we must use a, b, c (in some order) with possible substitutions. Therefore, the four cases considered earlier are the only ones.
But wait, is there a scenario where we use -a? The problem says that we can substitute -b or -c, but not -a. Therefore, a remains as +2.0 km east.
But suppose someone thought that maybe we can use -a. However, according to the problem statement, the original list includes a, and substituting is only for b and c. Therefore, a is fixed as +2.0 km east. Hence, the previous analysis holds.
So, the maximum so far is ~4.84 km. Now, check the options. Among the given options, option B is 4.6 km, H is 5.0 km, E is 5.2 km. Is there a higher possible?
Wait, perhaps I missed a combination where the three vectors are not including a. But the problem says that the three displacements are a, b, c. So you have to use all three. Therefore, we can't avoid a. So maybe the maximum is indeed ~4.84 km, which is closest to option H (5.0 km) or B (4.6 km). But why is there a discrepancy?
Alternatively, perhaps I'm missing a better configuration. Let me double-check calculations.
Another possibility: Could we arrange the vectors in a way that all components contribute positively to the east and north directions?
For example, if we use a, b, -c. As done in case 2, giving large east component and north component.
Alternatively, what if we use -b and -c in certain ways? Let me try another approach.
Suppose we use a, b, -c. Then as calculated, the displacement is (approx) 4.732 east, 1.0 north. Magnitude ~sqrt(4.732² + 1.0²) ≈ sqrt(22.39 + 1) ≈ sqrt(23.39) ≈ 4.837 km.
But is there a way to get even larger?
What if we use b and -c, but also adjust the order? Wait, but the order doesn't matter for summation. So the same thing.
Alternative idea: Suppose instead of using a, b, c, we use -a, b, c. But can we do that? Earlier conclusion was that we can't, since -a is not allowed. But if we could, that would be interesting. Let's explore this hypothetical case.
Assume that we can use -a. Then, the three vectors would be -a, b, c. Their components:
-a: (-2.0, 0)
b: (1.732, 1.0)
c: (-1.0, 0)
Summing: East: -2.0 +1.732 -1.0 = -1.268 km
North: 1.0 km
Magnitude: sqrt((-1.268)^2 + 1.0^2) ≈ sqrt(1.608 + 1) = sqrt(2.608) ≈ 1.615 km. Worse than previous cases.
Not helpful. Therefore, not useful.
Alternatively, using a, -b, -c. Which gave us (as per case 4):
East: 4.732, North: -1.0. So magnitude same as case 2.
Alternatively, what if we use a, -b, and then another vector? Wait, but the three vectors must be a, b, c (with possible substitutions). So the only options are the four cases already analyzed.
Alternatively, what if we use a, b, -c, but arrange the vectors differently? Doesn't help.
Alternatively, perhaps using other configurations where we have more contribution in the north-south direction. For example, if we use b twice? But we can't reuse vectors, as we have to use three distinct vectors. Wait, no, the problem says you may use the displacements in any order, but presumably, each displacement is used once. So we can’t repeat vectors. Hence, the three vectors must be a, b, c (with possible sign flips).
Another thought: What if we use a, -b, and then another vector. Wait, that’s covered in case 3 and 4.
Alternatively, what if we use b, c, and -c? That would be invalid since we can't use c and -c together in the three moves. Wait, no, the three moves are a, b, c (with possible substitutions). So if we use b, c, and -c, that's impossible since we can only use each displacement once. Hence, that's not allowed.
Wait, but the problem says "you may use the following displacements in any order: (a), (b), (c)". So you have to use each of the three once. So the three moves are a, b, c (each once), but you can choose to substitute either b with -b or c with -c. Therefore, the three vectors are a, ±b, ±c.
So, the four cases are the only possibilities.
So, returning to case 2 and 4, which yield ~4.84 km. Comparing to the options, the closest is option H (5.0 km), but 4.84 is less than 5.0. However, is there a way to achieve a greater magnitude?
Perhaps another combination. Let's think outside the box. What if we use a, b, and -b? No, that would duplicate b. Not allowed. Or a, -b, -b? Again, duplicates. Can't.
Wait, but the three moves must be three vectors. So you have to use each of a, b, c once, but possibly flipped.
Alternatively, maybe arranging the vectors in such a way that their contributions add constructively. For example, if we can get more east-west movement combined with north-south.
Wait, in case 2, the total east is 4.732, north is 1.0. What if we could increase the north component further?
How? By using a vector with more north component. But the only vector with significant north component is b, which is 2.0 km at 30°. Using -b reduces north component. So if we use b, the north component is 1.0 km. If we use -b, it's -1.0 km. So the maximum north component achievable is 1.0 km via b. Therefore, in case 2, the north component is maximized.
Alternatively, if we use both b and another vector contributing north. But since the other vectors are a (no north) and c (no north), the only source of north is b. Hence, the maximum north component is 1.0 km.
So, the north component is capped at 1.0 km. Therefore, the total displacement's magnitude depends on the east component and the north component. In case 2, the east component is 4.732 km, and north is 1.0 km. The resulting magnitude is sqrt(4.732² + 1²) ≈ 4.84 km.
Is there a way to have a larger east component? Yes. For example, if we can have more eastward movement. How?
By maximizing the east component. Let's see.
The east components come from:
- a: +2.0 km
- b: +2.0*cos(30°) ≈ 1.732 km
- c: -1.0 km
- -c: +1.0 km
So, if we use -c (instead of c), then the east component increases by 2.0 km (because originally, with c, it's +2.0 + 1.732 -1.0 = 2.732; with -c, it's +2.0 + 1.732 + 1.0 = 4.732). So that's what case 2 achieves.
Alternatively, what if we use -b and -c? Then:
East component would be a (2.0) + (-b)'s east component (still 1.732) + (-c)'s east component (1.0). Same as case 2: 4.732.
Alternatively, if we could somehow have more east component. For example, if we used -b and -c, but that's already accounted for.
Alternatively, if we could use -a, but that's not allowed.
Alternatively, what if we use a, -b, and -c? Which gives same east component as case 2.
So, the maximum east component is achieved in case 2 and 4, which is 4.732 km.
And the north component is 1.0 km.
Is there a way to have more north component? Let's think. If we could use two vectors contributing north, but the only one is b. So if we use b once, it contributes 1.0 km north. There's no other vector with north component besides b (and -b). Therefore, the maximum north component is 1.0 km.
Hence, the maximum magnitude is sqrt( (4.732)^2 + 1^2 ) ≈ 4.84 km. Now, comparing with the options, which are:
B: 4.6 km
H: 5.0 km
E: 5.2 km
Other options are lower or much smaller.
But 4.84 km is not listed. Wait, the options are:
Option B: 4.6 km
Option H: 5.0 km
Option E: 5.2 km
So, is there a mistake in the calculation? Let's redo the math more accurately.
Calculating 4.732^2:
Let me compute 4.732 × 4.732 precisely.
Breakdown:
First, write 4.732 as 4 + 0.732.
Square it: (4 + 0.732)^2 = 4² + 2×4×0.732 + 0.732² = 16 + 5.856 + 0.536 = ?
Wait:
4² = 16
2*4*0.732 = 8 * 0.732 = 5.856
0.732² = ?
Well, 0.7³ = 0.343, 0.73² = 0.5329, 0.732² ≈ (0.73 + 0.002)^2 = 0.73² + 2*0.73*0.002 + 0.002² ≈ 0.5329 + 0.00292 + 0.000004 ≈ 0.5358.
Adding all parts: 16 + 5.856 = 21.856 + 0.5358 ≈ 22.3918.
Add the square of the north component (1.0^2 = 1.0):
Total squared magnitude ≈ 22.3918 + 1.0 = 23.3918.
Square root of 23.3918 is approximately?
Well, 4.8^2 = 23.04
4.84^2 = (4.8 + 0.04)^2 = 4.8² + 2*4.8*0.04 + 0.04² = 23.04 + 0.384 + 0.0016 = 23.4256
Compare with 23.3918. So sqrt(23.3918) is slightly less than 4.84. Specifically, between 4.8 and 4.84.
Because 4.83^2 = ?
Calculate 4.83^2:
= (4.8 + 0.03)^2 = 4.8² + 2*4.8*0.03 + 0.03² = 23.04 + 0.288 + 0.0009 = 23.3289
Still less than 23.3918.
Next, 4.835^2:
Let me compute incrementally.
Between 4.83 and 4.84:
At x=4.83, x²=23.3289
At x=4.83 + d, x²=23.3289 + 2*4.83*d + d².
Set target 23.3918.
Difference: 23.3918 - 23.3289 = 0.0629.
Using linear approximation:
d ≈ 0.0629/(2*4.83) ≈ 0.0629 / 9.66 ≈ 0.0065.
So approx 4.83 + 0.0065 = 4.8365.
So sqrt(23.3918) ≈ 4.8365 km ≈ 4.84 km.
So the accurate value is approximately 4.84 km.
Given the options provided, the closest is option B (4.6 km), but that's significantly lower. Option H is 5.0 km. Why is that?
Maybe I made a mistake in assuming the vectors are added as a, b, -c. Let me verify the components again.
Wait, vector a is 2.0 km east (positive x-direction).
vector b is 2.0 km at 30° north of east, so components are 2.0*cos(30°) east, and 2.0*sin(30°) north.
vector -c is 1.0 km east (since c is west, -c is east).
So total east: 2.0 + 2.0*cos(30°) + 1.0.
Yes, that's 2.0 + 1.732 + 1.0 = 4.732 km east.
north: 2.0*sin(30°) = 1.0 km north.
That's correct.
So the total displacement is (4.732, 1.0). The magnitude is sqrt( (4.732)^2 + 1.0^2 ). As computed, about 4.84 km.
But the options don't have this value. The nearest option is H: 5.0 km. But why is that?
Alternatively, maybe there's a different combination where the magnitude is larger.
Wait, let's think again. What if instead of using a, b, -c, we use a different arrangement of vectors?
For example, suppose we use b, -c, and a. Wait, that's the same as case 2.
Alternatively, what if we use b, -c, and a, but in different directions? No, that's the same.
Wait, another idea: perhaps combining vectors in such a way that the resultant vector points purely in the east direction, thus maximizing the magnitude.
Can we achieve that?
To have zero north component. For example, if we use b and -b. But we can't use both b and -b in the three moves, since we have to use a, b, c (once each), but with possible substitutions. So if we use -b, we can't use b again.
Alternatively, using a, b, and -c. The north component is 1.0 km from b. If we could cancel that, but that requires another vector pointing south. But the only vector with southern component is -b. So if we use -b instead of b, then the north component is -1.0 km. But then, the east component would be 2.0 + 1.732 + 1.0 (if -c) or 2.0 + 1.732 -1.0 (if c). Wait, if we use -b and c, then:
case 3: a, -b, c.
Components:
East: 2.0 + 1.732 - 1.0 = 2.732 km.
North: 0 - 1.0 + 0 = -1.0 km.
Magnitude sqrt(2.732² + 1.0²) ≈ same as before, 2.91 km.
No improvement.
Alternatively, if we use a, -b, and -c: case 4, which gives east 4.732, north -1.0, same magnitude.
So no gain.
Alternatively, what if we use a, b, and -b? Impossible, since we can't use both b and -b.
Another thought: What if we use a, -b, and -c? As discussed, that gives same as case 2.
Alternatively, is there a way to have more than 1.0 km in north direction? For example, if we use both b and another vector with north component. But the only other vectors are a and c, which have no north component. So no.
Alternatively, if we use a, b, and -c, and then add another vector, but we can't, since we've already used three.
Wait, but maybe there's a different approach. Instead of thinking of the three vectors as a, ±b, ±c, maybe we can choose different substitutions for b and c.
Wait, for example, what if we use -a, -b, -c? But again, -a is not allowed.
Alternatively, what if we use a, -b, -c? Which is case 4, yielding the same result as case 2.
Hmm. I think I have exhausted all possible valid combinations.
Given that, the maximum possible is approximately 4.84 km, which is not explicitly listed in the options. However, looking closely at the options, option H is 5.0 km, and E is 5.2 km. These seem high, but perhaps there's a miscalculation.
Wait, let me recheck the arithmetic for the case where the east component is 4.732 and north is 1.0.
Wait, what if I made a mistake in calculating the east component? Let me recompute:
Vector a: 2.0 km east.
Vector b: 2.0 km at 30° north of east ⇒ east component is 2.0*cos(30°) = 2*(√3)/2 = √3 ≈ 1.732 km.
Vector -c: 1.0 km east.
Total east: 2.0 + 1.732 + 1.0 = 4.732 km.
Correct.
North component: only from vector b: 2.0*sin(30°) = 1.0 km.
So magnitude is sqrt( (4.732)^2 + 1.0^2 )
As previously calculated, ≈ 4.84 km.
But the options don't have this. However, looking at the options again, option E is 5.2 km, which is quite bigger. How is this possible?
Wait, is there a misinterpretation of the problem? Maybe the problem allows using more than three moves? No, the problem specifies three moves.
Alternatively, did I miss a scenario where the three vectors are not a, ±b, ±c, but something else?
Wait, the problem says: "you may use the following displacements in any order: (a)... (b)... (c)... Alternatively, you may substitute either −⃗b for ⃗b or −⃗c for ⃗c."
Does this mean that we can use any number of the vectors, as long as we substitute either -b or -c for b or c? Or does it mean that we have to use each of the three vectors once, with possible substitution of b or c?
The wording is ambiguous, but given that it's an orienteering class where you make three straight-line moves, it makes sense that you perform three moves, each being one of the vectors (with possible substitution). So, three moves, each being a, b, c, with possible substitution of b or c.
Therefore, the four cases are the only ones.
But then, why is option E (5.2 km) present? Let's think: what if the user chooses to use b and -c, but also another vector? No, three moves.
Alternatively, maybe I made a mistake in interpreting the angles. Let me check the components of vector b again.
Vector b is 2.0 km at 30 degrees north of east. So θ = 30° measured from the east axis towards north. So the components are:
East: |b| * cos(theta) = 2.0 * cos(30°) ≈ 1.732 km
North: |b| * sin(theta) = 2.0 * sin(30°) = 1.0 km.
Correct.
Another possibility: Did I mix up the sine and cosine? For example, sometimes people confuse the components. Let me confirm:
When a vector is at theta degrees north of east, the east component is adjacent side, hence cos(theta), and the north component is opposite side, hence sin(theta). Correct.
So that is correct.
Another thought: What if the three vectors are arranged such that the resultant is diagonal, but with longer legs? For example, if the vectors are arranged to go primarily east and north, but with a larger overall displacement.
But how?
Wait, perhaps using vector a, then vector b, then vector -c. As in case 2, giving 4.732 east and 1.0 north.
But what if we use vector b twice? But we can't, as we have to use three distinct vectors.
Wait, unless the problem allows repeating vectors. But the problem says "you may use the following displacements in any order", suggesting that each displacement is used once.
Alternatively, perhaps I misunderstood the problem. Maybe the three moves can be any combination of the vectors, not necessarily a, b, c. For example, you can use a, b, or c, or their negatives, but not necessarily all three.
Wait, but the problem says "you may use the following displacements in any order: (a)... (b)... (c)...". This usually implies that you have to use each of the three once. Otherwise, if you can use any subset, the problem would say something like "choose any combination of the displacements".
But given the ambiguity, perhaps the intended solution assumes that we can use any number of the vectors, including repeats, with substitutions. But that contradicts the usual understanding of such problems.
Alternatively, perhaps the problem allows using a single displacement repeated three times, but that seems unlikely.
Alternatively, maybe the three moves are allowed to be any of the vectors, with replacement, meaning you can use a, b, c repeatedly, but with possible substitution of b or c. For example, using a, b, b. But that would not align with standard problem conventions.
But if that is allowed, then we could potentially create a larger displacement. For example, using a, b, and b. But then, since you can substitute -b, perhaps using b and -b.
Wait, but this is too speculative. Without clear instructions, it's hard to know.
Alternatively, maybe the problem allows using any of the vectors (including their negatives), but not necessarily using all three. For example, you can use a, b, and then another vector, but that would require clarification.
But given the problem statement says "you may use the following displacements in any order: (a)... (b)... (c)...", it's more likely that you must use each of the three once. Hence, leading us back to the previous conclusion.
But since the options don't contain 4.84 km, perhaps the error lies elsewhere. Let me check the options again:
The options are labeled with (A)-(J). Let me check the numerical values again.
Wait, option H is 5.0 km, which is higher than our calculated 4.84. How could this happen?
Wait, what if the vectors are summed in a different way? Let me imagine a scenario where the east component is larger.
For example, using a, -b, and -c. Wait, no, that's the same as case 2.
Alternatively, what if we use vector b twice? But that's not allowed.
Alternatively, what if we use vector c and -c? But that would be two vectors, and then a third.
Wait, if we use a, c, -c, but that's invalid since you can't use c and -c in the same three moves. You have to use each once.
Alternatively, what if the three moves are a, -b, and -c, but the resultant vector is calculated incorrectly.
Wait, let's re-express the vectors numerically.
Let me compute the exact value of the magnitude.
Let me calculate 4.732^2 + 1^2.
Exact value of cos(30°) is √3/2 ≈ 0.8660254...
So vector b's east component is 2 * √3/2 = √3 ≈ 1.73205 km.
So total east component is 2.0 + √3 + 1.0 = 3.0 + √3 ≈ 3.0 + 1.73205 = 4.73205 km.
North component is 1.0 km.
So the magnitude is sqrt[(3.0 + √3)^2 + 1^2].
Expand (3 + √3)^2 = 9 + 6√3 + 3 = 12 + 6√3.
Adding 1^2: 12 + 6√3 + 1 = 13 + 6√3.
So the magnitude is sqrt(13 + 6√3).
Let me compute this value numerically.
First, √3 ≈ 1.73205.
So 6√3 ≈ 6 * 1.73205 ≈ 10.3923.
Adding 13 gives 13 + 10.3923 = 23.3923.
So sqrt(23.3923) ≈ 4.837 km, as before.
But what is sqrt(13 + 6√3) exactly? Let's compute this expression.
Alternatively, perhaps this equals 5.0 km? Let's check:
5.0^2 = 25. Our value is 23.39 < 25, so no.
Closest is 4.837 km, which is approximately 4.84 km.
But the options don't have this value. The next highest is option E: 5.2 km, which is higher. How could this occur?
Maybe the problem expects a different interpretation. Let me reconsider.
Perhaps the problem allows using the vectors in any order, and substituting either -b or -c for b or c, but not necessarily using all three. For example, using only two vectors, or even one. But that would violate the problem statement, which says "moving as far [...] as possible by making three straight-line moves."
So three moves, each being one of the vectors, possibly substituted.
Alternatively, perhaps I made a mistake in assuming that the vectors are a, b, c. Let me reread the problem:
"In an orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three straight-line moves. You may use the following displacements in any order: (a) a, 2.0 km due east (directly toward the east); (b) b, 2.0 km 30° north of east (at an angle of 30° toward the north from due east); (c) c, 1.0 km due west. Alternatively, you may substitute either −⃗b for ⃗b or −⃗c for ⃗c."
Ah! The phrase "you may use the following displacements in any order:..." suggests that you can choose any of the three vectors (a, b, c) in any order, and you can substitute either -b or -c for b or c. So, for example, you could choose to use a, -b, -c. Or a, -b, c. Or a, b, -c. Or a, b, c. But you must use three displacements, each being one of the three, with possible substitutes.
But perhaps the key insight is that you can choose any of the three vectors, with possible substitution, and you can choose the order. So, for example, you could choose to use vector a, vector -b, and vector -c, but that would be three displacements, and the total would be as calculated.
But the problem is asking for the maximum possible distance, so maybe there's a smarter combination.
Wait, what if we use vector -b and vector -c, but also vector a?
Wait, that's what we did in case 2 and 4.
Alternatively, what if we use vector b and vector -c, and then vector a. That's what we did.
But what if we use vector b, vector -c, and vector a, but in a different order? No difference.
Alternatively, what if we use vector -b, vector -c, and vector a? Same thing.
Another idea: What if we use vector a, vector b, and vector -b? But that would require using vector b and -b, which is not allowed since we have to use each vector once. So no.
Wait, but the problem says you may substitute either -b or -c for b or c. So if you choose to use b, you can choose to use it or -b. So, for example, you could use a, b, c, or a, -b, c, or a, b, -c, or a, -b, -c.
These are the four possible combinations.
But what if we use vector b and vector -b in the three moves? That would require using both b and -b, which is not allowed, since you can only use each vector once. Hence, no.
Therefore, the maximum is 4.84 km.
But this is not in the options. However, let's look at the options again. The options are:
(A): 4.2 m
(B): 4.6 km
(C): 3.8 km
(D): 2.9 km
(E): 5.2 km
(F): 3.5 m
(G): 4.8 m
(H): 5.0 km
(I): 5.6 m
(J): 3.2 km
Notice that option G is 4.8 m, which is 0.0048 km, which is much smaller than our calculated 4.84 km. Option H is 5.0 km, which is 0.5 km more than our calculation. But maybe the problem has a typo, or perhaps my calculation is wrong.
Alternatively, perhaps I made a mistake in the components. Let me try a different approach: using vector magnitudes and triangle inequality.
The maximum possible displacement occurs when all vectors are aligned in the same direction.
But with the constraints of the vectors, can we arrange them to point in the same direction?
For example, if we can have all vectors adding up to a single direction.
Let's try to maximize the east-west component.
If we use a, -b, -c, as in case 2, we get 4.732 km east.
If we use a, b, c, we get 2.732 east.
If we use a, -b, c, we get 2.732 east.
If we use a, b, -c, we get 4.732 east.
So the maximum east-west component is 4.732 km.
If we could also have a large north component, that would improve the magnitude.
But the maximum north component is 1.0 km (from b).
So the maximum magnitude is sqrt(4.732^2 + 1.0^2) ≈ 4.84 km.
But if the problem has a typo, and the correct answer is 4.8 km, which is option G, but it's in meters, which is inconsistent. Alternatively, if the calculation is wrong.
Wait, another possibility: the problem might expect the use of vector subtraction in a different manner. For example, if we use vector -b and -c, but in a different way.
Wait, let me try to visualize:
If we use vector a (east 2.0), vector -b (south-east 2.0 at 30°), and vector -c (east 1.0).
So the total displacement would be:
East: 2.0 + 2.0*cos(30°) + 1.0 = 4.732 km
South: 2.0*sin(30°) = 1.0 km.
But since it's south, the north component is -1.0 km.
So the magnitude is sqrt(4.732^2 + 1.0^2) = 4.84 km.
Same as before.
Alternatively, if we use vector a (east 2.0), vector b (northeast), and vector -c (east 1.0), the total is 4.732 east and 1.0 north.
So the same as before.
But what if we use vector a, vector b, and vector -b? No, that's not allowed.
Alternatively, what if we use vector a, vector b, and vector c, but in a different orientation?
No, the calculation is the same.
Given that, I think the correct answer is approximately 4.84 km, which is not listed. But since the options don't have this, perhaps I made a mistake in the calculation.
Wait, let's try to calculate the exact value of sqrt(13 + 6√3).
Let me compute 13 + 6√3 numerically:
√3 ≈ 1.73205080757
6√3 ≈ 10.3923048454
13 + 10.3923048454 ≈ 23.3923048454
Now, sqrt(23.3923048454) ≈ 4.837 km.
Now, looking at the options, the closest is option B: 4.6 km, which is 0.237 km less, or option H: 5.0 km, which is 0.163 km more.
But why is option H present? Maybe the problem expects a different approach.
Wait, another idea: what if the three vectors are not a, b, -c, but instead a, b, and -b? But that would require using b twice, which is not allowed.
Alternatively, perhaps the problem considers the possibility of using the vectors in a different way, like using vector -b and vector -c in such a way that they add up to a larger vector.
Wait, let's try to think of the three vectors as a, b, and -c, but what if we rearrange the order to add them in a way that their resultant is larger?
No, vector addition is associative and commutative, so the order doesn't matter.
Alternatively, maybe the problem expects the student to realize that the maximum is achieved by adding vectors in a specific way, like all vectors pointing in the same direction.
But given the vectors, is there a way to have all three vectors pointing in the same direction?
Vector a is east, vector b is northeast, vector c is west. If we use -c (east), and -b (southeast), but that would not be in the same direction.
Alternatively, if we use a, -b, -c, but that gives a mixture of east and south.
Alternatively, what if we use vector a, vector b, and vector -c, but arrange them so that the resultant is mainly east?
Yes, which is what we did.
But the magnitude is 4.84 km, which is closest to option B (4.6 km) or H (5.0 km). Since 4.84 is closer to 5.0 km than to 4.6 km, perhaps the intended answer is H.
But why would the calculation give 4.84 instead of 5.0? Maybe there's an error in the problem setup.
Alternatively, perhaps the problem expects us to use vector b and -c in a different way. Let me check if I got the angle right.
Vector b is 30° north of east. So the angle from the east is 30°, so the components are correctly calculated.
Alternatively, maybe the problem wants the maximum distance in a different unit? But the given displacements are in km, and the options are in km or meters, but the correct answer should be in km.
Alternatively, perhaps there's a calculation mistake in the components. Let me recompute the east and north components.
With a, b, -c:
East: 2.0 + 2.0*cos(30°) + 1.0 = 2.0 + 1.732 + 1.0 = 4.732 km.
North: 2.0*sin(30°) = 1.0 km.
Yes.
So, the magnitude is sqrt(4.732^2 + 1.0^2) ≈ 4.84 km.
But the options don't have this. However, looking at the options, option H is 5.0 km. Maybe the problem expects rounding up, or there is a miscalculation.
Alternatively, maybe the problem intends for the students to recognize that the maximum is achieved by adding all vectors in the same direction, but that's not possible with the given vectors.
Wait, what if we use vector a (east), vector b (northeast), and vector -c (east), and then add them in such a way that the resultant is diagonally east-north, but with larger magnitude.
But that's what we did, and it gives 4.84 km.
Alternatively, perhaps the problem is expecting to use vector b and -c in a way that their horizontal components add up to more.
Wait, vector b's east component is 1.732, vector -c is 1.0, and vector a is 2.0. Total 4.732.
If we could add more east-component vectors.
But no, since we have to use a, b, -c.
Alternatively, what if we use vector a, vector -b, and vector -c? That would give the same east component.
Wait, no. Let's recalculate:
If we use a (2.0 east), -b (which is 2.0 km at 30° south of east, so east component 1.732, south component 1.0), and -c (1.0 east).
Total east: 2.0 + 1.732 + 1.0 = 4.732.
Total north: 0 - 1.0 + 0 = -1.0.
Same as before.
So, the magnitude is same.
Thus, I conclude that the correct answer is approximately 4.84 km, which is closest to option H (5.0 km), despite the slight discrepancy.
But why is option H 5.0 km? Perhaps there's an error in the problem's options, or perhaps I missed a scenario.
Wait, let me try one more time. What if we use vector a, vector b, and vector c, but in a different way? No, that's case 1.
Alternatively, what if we use vector a, vector -b, and vector c? That's case 3, giving 2.732 east and -1.0 north, magnitude 2.91 km.
No.
Alternatively, what if we use vector a, vector -b, and vector -c? Case 4, giving 4.732 east and -1.0 north, magnitude 4.84 km.
Same as case 2.
Therefore, I think the answer should be 4.84 km, which is closest to option H (5.0 km). Even though it's not perfectly matched, perhaps the intended answer is H.
But why is option H 5.0 km? Let me check if there's a calculation error in assuming the vectors.
Wait, what if the angle is 30° from the north? No, the problem says 30° north of east.
Alternatively, what if the angle is measured from the y-axis? No, the problem says 30° north of east, so it's from the east.
Alternatively, perhaps the problem uses radians instead of degrees? Unlikely, since it specifies 30°.
Alternatively, let's assume that the angle is 30° from the north, but that would be 60° from the east. But the problem says 30° north of east.
I think the calculation is correct.
Therefore, given the options, the closest is option H: 5.0 km, although it's slightly higher than our calculation. Alternatively, perhaps the problem expects the use of Pythagorean theorem with rounded numbers.
Let me calculate 4.732^2 + 1.0^2 with more precision.
4.732^2 = (4 + 0.732)^2 = 16 + 2*4*0.732 + 0.732^2 = 16 + 5.856 + 0.535824 = 22.391824.
Plus 1.0^2 = 23.391824.
Now, sqrt(23.391824). Let's calculate this precisely.
We know that 4.8^2 = 23.04.
4.83^2 = 23.3289.
4.835^2 = (4.83 + 0.005)^2 = 4.83^2 + 2*4.83*0.005 + 0.005^2 = 23.3289 + 0.0483 + 0.000025 = 23.377225.
4.84^2 = 23.4256.
Target is 23.391824.
So between 4.835 and 4.84.
Let me interpolate.
From 4.835 (23.377225) to 4.84 (23.4256), the difference is 23.4256 - 23.377225 = 0.048375.
Our target is 23.391824 - 23.377225 = 0.014599.
Fraction = 0.014599 / 0.048375 ≈ 0.3017.
So approximate value is 4.835 + 0.3017 * 0.005 ≈ 4.835 + 0.0015 = 4.8365 km.
So approximately 4.8365 km, which is approximately 4.84 km.
Now, if the problem's options have a typo and the correct answer is 4.8 km (option G, but in meters, which is 4.8 m = 0.0048 km, which is way off), or if there's a misunderstanding in units.
Alternatively, perhaps the problem expects the answer to be 5.0 km, and the calculation is supposed to ignore the north component, but that would be incorrect.
Alternatively, maybe the problem is designed to use the maximum of the east component alone, ignoring the north component. But that would be wrong.
Alternatively, perhaps the problem is in error, and the correct answer is 4.8 km, but in the options, it's mistakenly listed as option G (4.8 m). But that would be a major inconsistency.
Alternatively, perhaps I made a mistake in the initial assumption that the three vectors are a, b, -c. Let me check if there's another combination.
Wait, what if the three vectors are b, -b, and -c? But that would require using b and -b, which is not allowed.
Alternatively, what if the three vectors are a, -b, and -c, but the resultant is calculated as follows:
East: 2.0 + 1.732 + 1.0 = 4.732.
North: 0 - 1.0 + 0 = -1.0.
Magnitude is sqrt(4.732^2 + 1.0^2) = 4.84 km.
Same as before.
Given that, and the absence of 4.84 km in the options, I think the problem might have a typo, or perhaps the correct answer is H: 5.0 km, and the expected answer is H.
Alternatively, perhaps the problem assumes that the vectors are added in a different way, like using vector a, vector b, and vector -c, but with a different calculation.
Wait, let's try to calculate the magnitude using exact expressions.
Earlier, we found the magnitude is sqrt(13 + 6√3).
Let me compute this value numerically with more precision.
√3 ≈ 1.7320508075688772
6√3 ≈ 10.392304845413263
13 + 10.392304845413263 = 23.392304845413263
Now, sqrt(23.392304845413263) ≈ 4.837 km.
This is approximately 4.84 km.
Now, let's compare with the options:
Option B: 4.6 km
Option H: 5.0 km
The difference between 4.84 and 4.6 is 0.24 km.
Between 4.84 and 5.0 is 0.16 km.
So 4.84 is closer to 5.0 km. Therefore, if forced to choose from the options, option H (5.0 km) is the closest.
But this is highly questionable, as 4.84 is closer to 5.0 than to 4.6. However, in real exams, sometimes the closest option is chosen, even if it's not exact.
Therefore, under this reasoning, the answer would be H: 5.0 km.
But I am unsure because the calculation shows it's 4.84, which is not in the options. However, given the options provided, H is the closest.
Alternatively, perhaps the problem expects using the vector b twice, but that is against the rules.
Alternatively, maybe the problem assumes that the angle is 30° from the north, leading to different components.
Let me check: if the vector b is 30° north of east, then the components are as before. If it were 30° from the north, it would be 60° from the east.
But the problem says 30° north of east, so it's 30° from the east towards north.
Therefore, I think the calculation is correct.
Given the options, and the fact that 4.84 is closer to 5.0 than to 4.6, the answer is H: 5.0 km.
**Final Answer**
\boxed{H}
</think>
### Understanding the Problem
We are tasked with determining **the maximum possible straight-line distance from base camp**, after performing **exactly three straight-line moves**, using the following sets of displacements:
- Displacement **$\vec{a}$**: 2.0 km due **east**
- Displacement **$\vec{b}$**: 2.0 km at **30° north of east**
- Displacement **$\vec{c}$**: 1.0 km due **west**
Additionally, we may **substitute** either:
- $-\vec{b}$ (i.e., 2.0 km at **30° south of east**)
- $-\vec{c}$ (i.e., 1.0 km due **east**)
---
### Strategy for Maximizing Distance
To maximize the straight-line distance from the starting point (base camp), we aim to **maximize the magnitude of the resultant vector** obtained from the sum of the three displacements. This involves computing the components (east and north) of each displacement vector, applying appropriate sign changes if required, and then computing the Euclidean norm of the total vector.
Key observations:
- Vectors follow **commutative addition** (order of application does not impact final result).
- The **maximum magnitude** happens when the components (especially east) are increased, and the north component is optimized.
- Substituting $-\vec{c}$ improves the eastward motion, whereas substituting $-\vec{b}$ introduces a downward (south) component.
---
### Best Configuration Analysis
After evaluating various combinations, the optimal configuration turns out to be:
#### Displacements Used:
- **$+\vec{a}$:** 2.0 km east
- **$+\vec{b}$:** 2.0 km at 30° north of east
- **$-\vec{c}$:** 1.0 km east
This configuration ensures:
- Maximum eastward displacement (**increased by using $-\vec{c}$**),
- Positive northward displacement (**from $\vec{b}$**),
- Avoidance of unnecessary cancellations.
#### Component Calculations:
##### From $+\vec{a}$:
- East: $+2.0$
- North: $0$
##### From $+\vec{b}$ ($2.0$ km at 30° north of east):
- East: $2.0 \cdot \cos(30^\circ) = 2.0 \cdot (\frac{\sqrt{3}}{2}) = \sqrt{3} \approx 1.732$
- North: $2.0 \cdot \sin(30^\circ) = 2.0 \cdot 0.5 = 1.0$
##### From $-\vec{c}$:
- East: $+1.0$
- North: $0$
#### Sum of Components:
- **East component:** $2.0 + 1.732 + 1.0 = 4.732$
- **North component:** $0 + 1.0 + 0 = 1.0$
#### Resultant Vector Magnitude:
$$
|\text{Result}| = \sqrt{(4.732)^2 + (1.0)^2}
\approx \sqrt{22.39 + 1.00} = \sqrt{23.39} \approx 4.84 \text{ km}
$$
---
### Matching With Options
Among the given options:
- **Option B**: 4.6 km ✅ Closer to 4.84 than 5.0?
- Difference: $4.84 - 4.6 = 0.24$
- **Option H**: 5.0 km ❌ Larger than 4.84
- Difference: $5.0 - 4.84 = 0.16$
While technically **closer to 5.0 km**, the true value is **approximately 4.84 km**, which is not directly listed.
However, if we interpret the intent behind the options, and **consider that the question provides a finite selection**, then **4.84 km is closest to 5.0 km**, especially since the other option (4.6 km) is farther away.
---
### Final Conclusion
Although the precise value is **not among the given options**, **option H (5.0 km)** is the **closest match** to the calculated value of **≈ 4.84 km**.
---
**Answer: H**
$$
\boxed{H}
$$